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DESIGN OF STAIR CASE (DESIGN OF LANDI

( EX 11.12 LIMIT STATE DESIGN ASHOK K JA

1 INPUT DATASINPUT Min Head room above steps 2.00 Mtr

INPUT Ht of Floor 3.20 Mtr LANDING SLAB BINPUT Raise of the Stair 0.15 Mtr

INPUT Tread of the Stair 0.25 Mtr WAIST SLAB

Going of the Stair 0.29 Mtr

No of Goings on Ht of Floor and Raise 21 Nos

INPUT No of Goings to be given 11 Nos Per flight = No of Going

The slope or Pitch of Stairway 27.3 Degree

Check for Raise and Going OK

INPUT Length of Landing A (2x) 1.5 Mtr

Total Length of Going G 3.19 Mtr

INPUT Length of Landing B (2y) 1.50 Mtr

INPUT Width of waist Slab 1.50 Mtr

Effective Span of Stair Flight 4.69 Mtr

INPUT Effective Span of Landing Slab A/B 3.30 Mtr

INPUT Width of Landing Slab A/B 1.50 Mtr

INPUT Grade of Concrete M 25

INPUT Grade of Steel Fe 500

Type of Support of Beam CON

= 26 for CAN/SS/CON '7/2

=1 1 =1 up to 10 m, L>10,

= 1.1 =Factor % Tension Rft = 1 =Factor of Compress

= 1 = Factor of Flanged B

Span / Eff depth Ratio 28.6

Min Effective Depth of Stair Slab = 120 mm

Adopt Overall Depth of Stair Slab D 200 mm 200

Effective Length of Waist Slab 4.81 Mtr

Width of Waist Slab B 1.50 Mtr

2 CALCULATION OF LOAD, BM & SF

Flight Load

Dead Load of step section 550 N/m DL of Step Section= 1/2

Dead Load of inclined Flight 1640 N/m DL of Inclined Slab = S

Super Imposed Dead Load 330 N/m Finishing Load = (R+G)

Dead Load per Sqm on Plan 8690 N/Sqm

INPUT Live Load on plan 5000 N/Sqm

CAN/SS/CON '7/20/26

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Total Load 13690 N/Sqm

Load Factor 1.5

Factored Load on Flight w 20535 N/Sqm

Total Factored Load on Flight 30.80 KN/m

Landing A

Dead Load of Landing 5000 N/Sqm

Super Imposed / Finish Load 750 N/Sqm

Total Dead Load 5750 N/Sqm

Live Load on plan 5000 N/Sqm


Load Factor 1.5

Factored Load on Landing A 16125 N/Sqm

Total Factored Load on Landing A 24.20 KN/m

Landing B






Load Factor 1.5

Factored Load on Landing B 16125 N/Sqm

Total Factored Load on Landing B 24.20 KN/m

Total Factored Load on Above length 12.94 KN/m

Reaction at Support B 71.64 KN

Reaction at Support A 76.71 KN

Point of Zero Shear from Support A 2.65 Mtr

Max BM on Waist Slab 106.41 KNm

Effective Length of Waist Slab 4.69 MtrMax SF on Waist Slab 76.71 KN

Total Factor Load on Landing Slab A 119.73 KN

Reaction at A from Two Flight 153.43 KN

Total Factor Load on Landing Slab A 273.16 KN

Dist 150mm from the wall and 75 mminside the wall only DL is considered

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Max BM in Landing slab A 112.68 KNm

Max SF in Landing slab A 136.58 KN

3 DESIGN OF MEMBER TO RESIST BENDING MOMENT

a Design of waist/ Flight Slab

Grade of Concrete M 25

Grade of Steel Fe 500

Width of Slab 1.5 Mtr Max Depth

Max BM Mx 106.41 KN-M fy

BM = (Const*fck) bd^2 3.318 bd^2 250

Calculated Eff Depth of Slab 157 mm 157 415

RESULT Adopt Effective Depth d 160 mm 500

INPUT Use Dia of Main rft 16 mm 550

Adopt Cover for Slab 30 mm Limiting Mo

RESULT Over all Depth of Slab D 200 mm

Width of Beam 1500 mmGrade of Concrete M 25 Concrete

Grade of Steel Fe 500 15

a= 0.87 *(fy^2/fck) 8700 20

b= -0.87 fy -435 25

c= m= Mu/(bd^2) 2.77 30

35

m= Mu/(bd^2) p %= (-b- sqrt(b^2-4ac))/2a At

2.77 0.749 1799 Sqmm

Min area of Tension Steel Ao=0.85*bd/fy 408.00 Sqmm

Max area of Tensile Steel = 0.04 bD 12000 Sqmm

Provide Area of Tension Steel 1799 Sqmm

Area of One Bar 201.14 Sqmm 16 mm Dia

RESULT No of Bars 9 Nos 16 mm Dia

RESULT Spacing of Distribution Bars 170 mm 16 mm Dia

Temp rft 0.15 % of gross area will be provided in the longitudinal direction300 Sqmm

INPUT Use 8 mm Dia bars



b Landing Slab A

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BM = (Const*fck) bd^2 3.318 bd^2 250

Calculated Eff Depth of Slab 150 mm 415

Adopt Effective Depth d 150 mm 500



Over all Depth of Slab D 190 mm

Width of Slab 1500 mm

Grade of Concrete M 25 Concrete


a= 0.87 *(fy^2/fck) 8700 20

b= -0.87 fy -435 25

c= m= Mu/(bd^2) m= Mu/(bd^2) 30

35


3.32 0.939 2120 Sqmm

Min area of Tension Steel Ao=bd/fy 451.38 Sqmm






Temp rft 0.15 % of gross area will be provided in the longitudinal direction

285 Sqmm




4 DESIGN/ CHECK FOR MEMBER TO RESIST SHEAR

a Check for Shear in Waist Slab


Eff Depth of Slab 160 mm

Over all Depth of Slab 200 mm Grade of Con

Width of Slab 1500 mm Max SS N/

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Dia of Main rft 16 mm

Area of One Bar 201.14 Sqmm

No of Bars 9 Nos

Max Shear Force wl/2 76.71 KN

Percentage of Tensile Steel 100At/2bd = 0.75 %

Grade of

Design Shear Strength 0.575 N/ Sqmm Max SS N/

Calculated k Value 1.15

INPUT For 200 mm thick slab, k= 1.15

Permissible Max Shear Stress 0.661 N/ Sqmm

Nominal Shear stress Vu/bd 0.32 N/ Sqmm

Maximum Shear stress Tcm 3.10 N/ Sqmm

Shear Check Safe

Design of Stirrups



Effective Depth of Slab 160 mm

Over all Depth of Beam 200 mm

Width of Beam 1500 mmMax Shear Force wl/2 Vu 76.71 KN

Strength of Shear rft Vus=Vu-Tc bd 0 N Stirrup Rft NOT REQU

INPUT Dia of Shear rft 8 mm


No of legged vertical stirrups 2 Nos

Area of Vertical Stirrup Rft Asv 100.57 mm

Spacing of Shear rft x=0.87 fy Asv d/ Vus 0 mm

Check for Spacing NOT OK Min Spacing is 100 mm

Min Area of Shear rft 0.4 b x /fy 0 SqmmCheck for min Shear rft Area NOT OK

b Check for Shear in Landing Slab A


Effective Depth of Beam 150 mm

Over all Depth of Beam 190 mm Grade of Con

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Width of Beam 1500 mm Max SS N/



No of Bars 11 Nos



Grade of







Shear Check Un safe

Design of Stirrups




Over all Depth of Beam 190 mmWidth of Beam 1500 mm

Max Shear Force wl/2 Vu 76.71 KN

Strength of Shear rft Vus=Vu-Tc bd 6845 N



INPUT No of legged vertical stirrups 2 Nos


RESULT Spacing of Shear rft x=0.87 fy Asv d/ Vus 2160 mm 420 12

INPUT Provide Spacing of Shear Rft 180 180

Check for Spacing OK Min Spacing is 100 mm

Min Area of Shear rft 0.4 b x /fy 216 Sqmm

Check for Min Shear rft Area OK If NOT OK then Increa

5 CKECK FOR DEVELOPMENT LENGTH

a At Long Edge

Max Shear Force wl/2 136.58 KN Design Bo

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Grade of Concrete M 25 Greade 15

Grade of Steel Fe 500 Tbd N/Sqmm 1.0

Adopt Effective Depth d 160 mm

Dia of Slab rft 16 mm

Area of Tension rft 1799 Sqmm


Spacing of Bars 110 mm

Assumed Development Length 200 mm

INPUT Bond Stress Tbd 1.6 N/Sqmm

Development Length based on Anchorage Bond

Ld= 0'0.87 fy / 4Tbd

Ld in Anchorage Bond 1088 mm

Development Length based on Flexural Bond

Ld= 1.3 M1/V + Lo

INPUT Assumed Development Length 188 mm

Moment of Resistance offered by 16 mm dia bar @

M1 = 56,361,000 Nmm

V = 136578.90 N Developm

Ld in Flexural Bond 724 mm fy N/Sqmm

Factor of Develop Length 56 M15

Develop length of Single Bar 896 mm 250 55

415 56Max Development Length 1088 mm 500 69

Max Bar Size in Develop Length 16 mm

Check for Development Length OK

CKECK FOR DEVELOPMENT LENGTH

b At Short Edge



Grade of Steel Fe 500 Tbd N/Sqmm 1.0Adopt Effective Depth d 160.00 mm

Width of Beam b 1500 mm




No of Bars 9 mm

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Ld= 0'0.87 fy / 4Tbd



Ld= 1.3 M1/V + Lo



M1 = 106,439,000 Nmm


Ld in Flexural Bond 1201 mm fy N/SqmmFactor of Develop Length 56 M15


415 56

Max Development Length 1201 mm 500 69


Check for Development Length OK IfNOT OK Then Reduc

6 CHECK FOR DEFLECTION

Short Span of Slab 4.7 MtrEffective Depth d 160 mm




Area of One Bar 201 Sqmm


Percentage of tension steel at Mid Span = 0.75 %

= 26 for CAN/SS/CON '7/2=1 1 =1 up to 10 m, L>10,

INPUT Cal = 1.02 = 1.1 =Factor % Tension Rft

= 1 =Factor of Compress


Allowable L/d 28.6

Actual L/d 29.31

Tensi

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Deflection Check is NOT OK

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NG SLAB A AND STAIR FLIGHT)

IN)

190 mm Th with RFT 16 mm dia @ 140 C/C 11 Nos


/ ( 2 or 3)

1.5 3.19 1.50

3.30

2X GOING( G) 2Y

0/26

/10

=1 for 1 %, Fig 10.1ion Rft 1 for 0 % Fig 10.2

am 1 for web L=B Fig 10.3

8 dia @ 170 C/C

200 th with rft 16 dia @ 170 C/C


* Going * Raise * 25000 8 dia @ 180 C/C

qrt( G^2+R^2) *t * 25000

Finish thick * 25000

LANDING A LANDING B

STAIR FLIGHT

STAIR FLIGHT

BEAM

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of Nutral Axis

Xm=0.0035/(.0055+0.87*fy/Es), Es= 200000 N/Sqmm

0.53 d

0.48 d

0.46 d

0.44 d

ment of resistance MR = Const * b*d^2 N mm

Const= 0.36*fck*Xm(1-0.42*Xm)

Fe 250 Fe 415 Fe 500 Fe 550

2.229 2.067 1.991 1.949

2.972 2.755 2.655 2.598

3.715 3.444 3.318 3.248

4.458 4.133 3.982 3.897

5.201 4.822 4.645 4.547

Steel

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of Nutral Axis

Xm=0.0035/(.0055+0.87*fy/Es), Es= 200000 N/Sqmm

0.53 d

0.48 d

0.46 d

0.44 d



Fe 250 Fe 415 Fe 500 Fe 550

2.229 2.067 1.991 1.949

2.972 2.755 2.655 2.598

3.715 3.444 3.318 3.248

4.458 4.133 3.982 3.897

5.201 4.822 4.645 4.547

Max Shear Stress

crete M 25

qmm 3.1

Steel

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fck 25

Design Shear Strength

100 As bd SS N/Sqmm

0.75 3.87 0.575

Max Shear Stress

oncrete M 15 20 25 30 35

qmm 2.5 2.8 3.1 3.5 3.7

Value of K

Ds >300 275 250 225 200 175

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qmm 3.1

fck 25


100 As bd SS N/Sqmm

0.13 22.99 0.269

Max Shear Stress

oncrete M 15 20 25 30 35

qmm 2.5 2.8 3.1 3.5 3.7

Value of K

Ds >300 275 250 225 200 175

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20 25 30 35 40

1.2 1.4 1.5 1.7 1.9

220 mm C/C

nt Length for Single Bars

M20 M25 M30 M40 M15 M20 M25 M30 M40

46 44 37

47 40 38 30 45 38 32 31 2458 49 46 36 54 46 39 36 29

nd Stress

20 25 30 35 40

1.2 1.4 1.5 1.7 1.9

Tension Bars Compression Bars

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18 mm C/C


M20 M15 M20

46 44 37

47 45 38

58 54 46

e the Spacing of Bars

0/26/10

=1 for 1 %, Fig 10.1

ion Rft 1 for 0 % Fig 10.2


n Bars Compression Bars

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(Distribution Bars)

(Distribution Bars)

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40

4.0

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40

4.0

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DESIGN OF STAIR CASE W/O EDGE BEAM (

( EX 11.12 LIMIT STATE DESIGN ASHOK K JA


INPUT Ht of Floor 3.20 Mtr LANDING SLAB BINPUT Raise of the Stair 0.15 Mtr

INPUT Tread of the Stair 0.25 Mtr WAIST SLAB



INPUT No of Goings to be given 11 Nos Per flight = No of Going






INPUT Width of waist Slab 1.50 Mtr

Effective Span of Stair Flight 4.69 Mtr

INPUT Effective Span of Landing Slab A 3.30 Mtr

INPUT Width of Landing Slab A 1.50 Mtr





=1 1 =1 up to 10 m, L>10,

= 1.1 =Factor % Tension Rft = 1 =Factor of Compress



Min Effective Depth of Beam = 120 mm

Adopt Overall Depth of Beam D 200 mm 200

Effective Length of Waist Beam 4.81 Mtr

Width of Waist Slab B 1.50 Mtr


Flight Load

Dead Load of step section 550 N/m DL of Step Section= 1/2

Dead Load of inclined Flight 1640 N/m DL of Inclined Slab = S

Super Imposed Dead Load 330 N/m Finishing Load = (R+G)

Dead Load per Sqm on Plan 8690 N/Sqm


CAN/SS/CON '7/20/26

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Load Factor 1.5



Landing A






Load Factor 1.5

Factored Load on Landing A 16125 N/Sqm

Total Factored Load on Landing A 24.20 KN/m

Landing B






Load Factor 1.5

Factored Load on Landing B 16125 N/Sqm

Total Factored Load on Landing B 24.20 KN/m

Total Factored Load on Above length 12.94 KN/m

Reaction at Support B 71.64 KN

Reaction at Support A 76.71 KN

Point of Zero Shear from Support A 2.65 Mtr

Max BM on Waist Slab 106.41 KNm 110.09 FACTORE

Effective Length of Waist Slab 4.69 MtrMax SF on Waist Slab 76.71 KN 73.77 FACTORE

APROX M

Total Factor Load on Landing Slab A 119.73 KN 119.79 FACTORE

Reaction at A from Two Flight 153.43 KN 162.30 REACTIO

Total Factor Load on Landing Slab A 273.16 KN 282.09

Dist 150mm from the wall and 75 mminside the wall only DL is considered

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Max BM in Landing slab A 116.36 KNm TOTAL FA

Max SF in Landing slab A 141.05 KN TOTAL FA







BM = (Const*fck) bd^2 3.318 bd^2 250






Width of Beam 1500 mmGrade of Concrete M 25 Concrete


a= 0.87 *(fy^2/fck) 8700 20

b= -0.87 fy -435 25

c= m= Mu/(bd^2) 2.87 30

35


2.87 0.781 1875 Sqmm







Temp rft 0.15 % of gross area will be provided in the longitudinal direction300 Sqmm




b Landing Slab A

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BM = (Const*fck) bd^2 3.318 bd^2 250

Calculated Eff Depth of Slab 153 mm 415




Over all Depth of Slab D 190 mm

Width of Slab 1500 mm



a= 0.87 *(fy^2/fck) 8700 20

b= -0.87 fy -435 25

c= m= Mu/(bd^2) m= Mu/(bd^2) 30

35


3.32 0.939 2155 Sqmm








285 Sqmm





a Check for Shear in Waist Slab





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No of Bars 10 Nos



Grade of







Shear Check Safe

Design of Stirrups






Strength of Shear rft Vus=Vu-Tc bd 0 N Stirrup Rft NOT REQU






Check for Spacing NOT OK Min Spacing is 100 mm

Min Area of Shear rft 0.4 b x /fy 0 SqmmCheck for min Shear rft Area NOT OK

b Check for Shear in Landing Slab A




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No of Bars 11 Nos



Grade of







Shear Check Un safe

Design of Stirrups




Over all Depth of Beam 190 mmWidth of Beam 1500 mm

Max Shear Force wl/2 Vu 73.77 KN




INPUT No of legged vertical stirrups 2 Nos


RESULT Spacing of Shear rft x=0.87 fy Asv d/ Vus 4580 mm 420 12

INPUT Provide Spacing of Shear Rft 180 180


Min Area of Shear rft 0.4 b x /fy 216 Sqmm

Check for Min Shear rft Area OK If NOT OK then Increa


a At Long Edge


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Ld= 0'0.87 fy / 4Tbd



Ld= 1.3 M1/V + Lo



M1 = 56,361,000 Nmm





415 56Max Development Length 1088 mm 500 69




b At Short Edge



Grade of Steel Fe 500 Tbd N/Sqmm 1.0Adopt Effective Depth d 160.00 mm





No of Bars 10 mm

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Ld= 0'0.87 fy / 4Tbd



Ld= 1.3 M1/V + Lo



M1 = 110,109,000 Nmm


Ld in Flexural Bond 1203 mm fy N/SqmmFactor of Develop Length 56 M15


415 56





Short Span of Slab 4.7 MtrEffective Depth d 160 mm







= 26 for CAN/SS/CON '7/2=1 1 =1 up to 10 m, L>10,

INPUT Cal = 1.02 = 1.1 =Factor % Tension Rft



Allowable L/d 28.6

Actual L/d 29.31

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Deflection Check is NOT OK

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ESIGN OF LANDING SLAB A AND STAIR FLIGHT)

IN)



/ ( 2 or 3)

1.5 3.19 1.50

3.30

2X GOING( G) 2Y

0/26

/10



8 dia @ 170 C/C



* Going * Raise * 25000 8 dia @ 180 C/C

qrt( G^2+R^2) *t * 25000

Finish thick * 25000

LANDING A LANDING B

STAIR FLIGHT

STAIR FLIGHT

LANDING A LANDING B

STAIR FLIGHT

STAIR FLIGHT

BEAM

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LOAD * WAIST LENGTH / 8 * FACTOR 1.3

LOAD * WAIST LENGTH /2 * FACTOR 1.3

THOD OF CALCULATION

LOAD * LENGTH * WIDTH OF LAANDING SLAB A

FROM WAIST SLAB * 2 NOS * FACTOR 1.1

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TORED LOAD * LENGTH/ 8

TORED LOAD/2

of Nutral Axis

Xm=0.0035/(.0055+0.87*fy/Es), Es= 200000 N/Sqmm

0.53 d

0.48 d

0.46 d

0.44 d



Fe 250 Fe 415 Fe 500 Fe 550

2.229 2.067 1.991 1.949

2.972 2.755 2.655 2.598

3.715 3.444 3.318 3.248

4.458 4.133 3.982 3.897

5.201 4.822 4.645 4.547

Steel

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of Nutral Axis

Xm=0.0035/(.0055+0.87*fy/Es), Es= 200000 N/Sqmm

0.53 d

0.48 d

0.46 d

0.44 d



Fe 250 Fe 415 Fe 500 Fe 550

2.229 2.067 1.991 1.949

2.972 2.755 2.655 2.598

3.715 3.444 3.318 3.248

4.458 4.133 3.982 3.897

5.201 4.822 4.645 4.547

Max Shear Stress

crete M 25

qmm 3.1

Steel

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fck 25


100 As bd SS N/Sqmm

0.78 3.72 0.584

Max Shear Stress

oncrete M 15 20 25 30 35

qmm 2.5 2.8 3.1 3.5 3.7

Value of K

Ds >300 275 250 225 200 175

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qmm 3.1

fck 25


100 As bd SS N/Sqmm

0.12 23.36 0.267

Max Shear Stress

oncrete M 15 20 25 30 35

qmm 2.5 2.8 3.1 3.5 3.7

Value of K

Ds >300 275 250 225 200 175

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20 25 30 35 40

1.2 1.4 1.5 1.7 1.9

220 mm C/C


M20 M25 M30 M40 M15 M20 M25 M30 M40

46 44 37

47 40 38 30 45 38 32 31 2458 49 46 36 54 46 39 36 29

nd Stress

20 25 30 35 40

1.2 1.4 1.5 1.7 1.9


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20 mm C/C


M20 M25 M30 M40 M15 M20 M25 M30 M40

46 44 37

47 40 38 30 45 38 32 31 24

58 49 46 36 54 46 39 36 29


0/26/10

=1 for 1 %, Fig 10.1




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(Distribution Bars)

(Distribution Bars)

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40

4.0

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40

4.0

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DESIGN OF STAIR CASE WITH CENTRE BEA

( EX 7.5 RCC PUNMIA)


INPUT Ht of Floor 2.10 MtrINPUT Raise of the Stair 0.15 Mtr

INPUT Tread of the Stair 0.25 Mtr



INPUT No of Goings to be given 14 Nos






INPUT Width of Stair Case 1.50 Mtr

Effective Span for Stair 7.06 Mtr

INPUT Effective Span of Landing Slab 3.15 Mtr

INPUT Width of Landing Slab 1.50 Mtr

INPUT Provided Thickness of Beam 0.25 Mtr





=1 1 =1 up to 10 m, L>10,

= 1.1 CAL VALUE =Factor % Tension Rft




Min Effective Depth of Beam = 110 mm

Adopt Overall Depth of Beam D 450 mm 450

Effective Length of Waist Beam 7.17 Mtr

Width of Beam 0.25 Mtr


Dead Load of step section 550 N/m

Dead Load of inclined Flight 3680 N/m

Super Imposed Dead Load 330 N/m

Dead Load on Plan 5140 N/Sqm



CAN/SS/CON '7/20/26

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Load Factor 1.5



Cantilever Span of Step 0.63 Mtr

Cantilever Moment = W*L 4.45 KNm


a Design of Cantilever Slab Pr to Centre Beam





BM = (Const*fck) bd^2 3.444 bd^2 250


RESULT Adopt Effective Depth d 80 mm 500INPUT Use Dia of Main rft 8 mm 550



Width of Beam 1000 mm



a= 0.87 *(fy^2/fck) 5993.43 20

b= -0.87 fy -361.05 25

c= m= Mu/(bd^2) 0.70 30

35m= Mu/(bd^2) p %= (-b- sqrt(b^2-4ac))/2a At

0.70 0.199 160 Sqmm








165 Sqmm



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Effective Span L 7.17

Max BM in Center Beam W*L^2/8 146.52 KNm

Max SF in Centre Beam W*L/2 81.74 KN

b Design of Beam at Centre



Width of Beam 0.25 Mtr Max Depth


BM = (Const*fck) bd^2 3.444 bd^2 250

Calculated Eff Depth of Beam 412 mm 200 415


INPUT Use Dia of Main rft 16 mm 550Adopt Cover for Beam 30 mm Limiting Mo

Over all Depth of Beam D 450 mm




a= 0.87 *(fy^2/fck) 5993.43 20

b= -0.87 fy -361.05 25

c= m= Mu/(bd^2) m= Mu/(bd^2) 30

35


3.44 1.188 1226 Sqmm







a Check for Shear in Centre Beam


Eff Depth of Beam 412 mm



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No of Bars 7 Nos



Grade of







Shear Check Un safe

Design of Stirrups










Area of Vertical Stirrup Rft Asv 101 mm 96



Min Area of Shear rft 0.4 b x /fy 77.11 SqmmCheck for min Shear rft Area OK


a At Long Edge



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Ld= 0'0.87 fy / 4Tbd



Ld= 1.3 M1/V + LoINPUT Assumed Development Length 124 mm


M1 = 12,041,000 Nmm





415 56

Max Development Length 451 mm 500 69Max Bar Size in Develop Length 8 mm



b At Short Edge




Adopt Effective Depth d 412 mmWidth of Beam b 1000 mm




No of Bars 4 mm



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Ld= 0'0.87 fy / 4Tbd



Ld= 1.3 M1/V + Lo



M1 = 24,242,000 Nmm





415 56





Span of Beam 7.1 Mtr

Effective Depth d 412 mm

Width of Beam b 250 mmDia of Slab rft 8 mm






=1 1 =1 up to 10 m, L>10,

INPUT Cal = 1.18 = 1.1 =Factor % Tension Rft = 1 =Factor of Compress


Allowable L/d 28.6

Actual L/d 17.12

Deflection Check is OK

Tensi

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M

0/26

/10

=1 for 1 %, Fig 10.1



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of Nutral Axis

Xm=0.0035/(.0055+0.87*fy/Es), Es= 200000 N/Sqmm

0.53 d

0.48 d

0.46 d0.44 d



Fe 250 Fe 415 Fe 500 Fe 550

2.229 2.067 1.991 1.949

2.972 2.755 2.655 2.598

3.715 3.444 3.318 3.248

4.458 4.133 3.982 3.897

5.201 4.822 4.645 4.547

Steel

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of Nutral Axis

Xm=0.0035/(.0055+0.87*fy/Es), Es= 200000 N/Sqmm

0.53 d

0.48 d

0.46 d

0.44 dment of resistance MR = Const * b*d^2 N mm


Fe 250 Fe 415 Fe 500 Fe 550

2.229 2.067 1.991 1.949

2.972 2.755 2.655 2.598

3.715 3.444 3.318 3.248

4.458 4.133 3.982 3.897

5.201 4.822 4.645 4.547

Max Shear Stress

crete M 25

qmm 3.1

Steel

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fck 25


100 As bd SS N/Sqmm

0.16 18.27 0.299

Max Shear Stress

oncrete M 15 20 25 30 35

qmm 2.5 2.8 3.1 3.5 3.7

Value of K

Ds >300 275 250 225 200 175

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1.2 1.4 1.5 1.7 1.9

620 mm C/C


M20 M25 M30 M40 M15 M20 M25 M30 M40

46 44 37

47 40 38 30 45 38 32 31 24

58 49 46 36 54 46 39 36 29

nd Stress

20 25 30 35 40

1.2 1.4 1.5 1.7 1.9

Compression BarsTension Bars

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8 mm C/C


M20 M15 M20

46 44 37

47 45 38

58 54 46


0/26

/10



n Bars Compression Bars

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40

4.0

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DESIGN OF CANTILEVER STAIR CASE ONE

1 INPUT DATAS (EX 6.4 DHAYARATNAM)

INPUT Width of Cantilever Slab L 1.20 Mtr

INPUT Assume Thickness of Step L/8 0.15 MtrINPUT Assume Tread b 0.35 Mtr

CALCULATION OF LOAD, BM & SF

Self Wt of Step 1.6 KN/m

Finishes 0.4 KN/m

Live Load 1.5 KN/m

Total Load of one step 3.5 KN/m

Factor of Safety 1.00 KN/m

Design Load w 3.50 KN/m

Max BM=w*L/2 2.1 KN/m

Max SF = w*L 4.2 KNINPUT Grade of Concrete M 25








BM = (Const*fck) bd^2 3.444 bd^2 250









a= 0.87 *(fy^2/fck) 5993.43 20

b= -0.87 fy -361.05 25

c= m= Mu/(bd^2) 0.94 3035


0.94 0.272 77 Sqmm

Min area of Tension Steel Ao=0.12*bd 144.00 Sqmm


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180 Sqmm





a Check for Shear in Waist SlabGrade of Concrete M 25






No of Bars 2 Nos



Grade of







Shear Check Safe


b At Short Edge


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DGE OF STEP IS FIXED ON WALL

of Nutral Axis

Xm=0.0035/(.0055+0.87*fy/Es), Es= 200000 N/Sqmm

0.53 d

0.48 d

0.46 d

0.44 d



Fe 250 Fe 415 Fe 500 Fe 550

2.229 2.067 1.991 1.949

2.972 2.755 2.655 2.598

3.715 3.444 3.318 3.248

4.458 4.133 3.982 3.8975.201 4.822 4.645 4.547

Steel

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Max Shear Stress

crete M 25

qmm 3.1

fck 25


100 As bd SS N/Sqmm

0.51 5.64 0.495

Max Shear Stress

oncrete M 15 20 25 30 35

qmm 2.5 2.8 3.1 3.5 3.7

Value of K

Ds >300 275 250 225 200 175

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20 25 30 35 40

1.2 1.4 1.5 1.7 1.9

4 mm C/C


M20 M25 M30 M40 M15 M20 M25 M30 M40

46 44 3747 40 38 30 45 38 32 31 24

58 49 46 36 54 46 39 36 29


Compression BarsTension Bars

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40

4.0

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03. Staircase (1)